Problem: There are some tasks which have the following structure: you are given a model, and you can do some operations, you should use these operations to achive the goal. One way to create a new task is to use the same model and same operations, but change the goal.

Let's have a try. I have created the following task for Topcoder SRM 557 Div1-Hard: you are given n integers x1, x2, ..., xn. You are allowed to perform the assignments (as many as you want) of the following form xi ^= xj (in the original task i and j must be different, but in this task we allow i to equal j). The goal is to maximize the sum of all xi.

Now we just change the goal. You are also given n integers y1, y2, ..., yn. You should make x1, x2, ..., xn exactly equal to y1, y2, ..., yn. In other words, for each i number xi should be equal to yi.

Input

The first line contains an integer n (1 ≤ n ≤ 10000). The second line contains n integers: x1 to xn (0 ≤ xi ≤ 109). The third line contains n integers: y1 to yn (0 ≤ yi ≤ 109).

Output

If there is no solution, output -1.

If there is a solution, then in the first line output an integer m (0 ≤ m ≤ 1000000) – the number of assignments you need to perform. Then print m lines, each line should contain two integers i and j (1 ≤ i, j ≤ n), which denote assignment xi ^= xj.

If there are multiple solutions you can print any of them. We can prove that under these constraints if there exists a solution then there always exists a solution with no more than 106 operations.

Examples

Input

2
3 5
6 0


Output

2
1 2
2 2


Input

5
0 0 0 0 0
1 2 3 4 5


Output

-1


Input

3
4 5 6
1 2 3


Output

5
3 1
1 2
2 2
2 3
3 1


Input

3
1 2 3
4 5 6


Output

-1

Note

Assignment a ^= b denotes assignment a = a ^ b, where operation "^" is bitwise XOR of two integers.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
inline int read() {
  int x = 0;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  return x;
}
vector<pair<int, int> > ans1, ans2;
int x[10010], y[10010];
int n;
void solve() {
  int i, j;
  i = 1, j = 30;
  while (i <= n && j >= 0) {
    for (int k = i; k <= n; k++) {
      if (y[k] & (1 << j)) {
        if (i == k) break;
        y[i] ^= y[k];
        ans1.push_back(make_pair(i, k));
        break;
      }
    }
    if ((y[i] & (1 << j)) == 0) {
      j--;
      continue;
    }
    for (int k = 1; k <= n; k++) {
      if (k != i && (y[k] & (1 << j))) {
        y[k] ^= y[i];
        ans1.push_back(make_pair(k, i));
      }
    }
    i++;
    j--;
  }
  i = 1, j = 30;
  while (i <= n && j >= 0) {
    for (int k = i; k <= n; k++) {
      if (x[k] & (1 << j)) {
        if (i == k) break;
        x[i] ^= x[k];
        ans2.push_back(make_pair(i, k));
        break;
      }
    }
    if ((x[i] & (1 << j)) == 0) {
      j--;
      continue;
    }
    for (int k = 1; k <= n; k++) {
      if (k != i && (x[k] & (1 << j)) != (y[k] & (1 << j))) {
        x[k] ^= x[i];
        ans2.push_back(make_pair(k, i));
      }
    }
    if ((y[i] & (1 << j)) == 0) {
      x[i] = 0;
      ans2.push_back(make_pair(i, i));
      j--;
      continue;
    }
    i++;
    j--;
  }
}
int main() {
  n = read();
  for (int i = 1; i <= n; i++) x[i] = read();
  for (int i = 1; i <= n; i++) y[i] = read();
  solve();
  for (int i = 1; i <= n; i++) {
    if (x[i] != y[i]) {
      puts("-1");
      return 0;
    }
  }
  printf("%d\n", ans1.size() + ans2.size());
  for (int i = 0; i < ans2.size(); i++)
    printf("%d %d\n", ans2[i].first, ans2[i].second);
  for (int i = ans1.size() - 1; i >= 0; i--)
    printf("%d %d\n", ans1[i].first, ans1[i].second);
  return 0;
}